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Separation of variables

  1. Jul 5, 2009 #1
    this may seem like a simple question but how does one know that separation of variables for solving linear PDE's will work. What i mean is that it seems to pick out a form of the solution to a given problem (I have heard that linear PDE's have an infinite number of functions of a particular form, e.g. for the wave equation the solution is of the form f(x-vt) + g(x+vt)). I can understand that for problems in e&m the separation of variables technique picks out a particular form (like for a cartesian coordinates for laplace's equation for a box, the solutions come out to be sines and cosines), but what about linear PDE's in finance (like the Black scholes equation). Thanks in advance to anyone who can clarify this.
  2. jcsd
  3. Jul 5, 2009 #2


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    Didn't we just have this? Or was that on another forum?

    Do you agree that if F(x)= G(x), then [itex]\int F(x)dx= \int G(x)dx[/itex]?
    (If not I have no idea what to say!)

    If [itex]dy/dx= f(x)g(y)[/itex] then [itex](1/g(y))dy/dx= f(x)[/itex]. Since y is itself a function of x, this is the "F(x)= G(x)" above.

    Then [itex]\int [(1/g(y)) dy/dx]dx= \int f(x)dx[/itex]. And, of course, (dy/dx)dx= dy so this is [itex]\int (1/g(y))dy= \int f(x)dx[/itex].
  4. Jul 5, 2009 #3
    I was actually talking about partial differential equations like Laplace's equation or other such linear equations
  5. Jul 5, 2009 #4


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    Yes PDE's have families of solutions, it's the boundary and initial conditions that let you pin down the actual solution. It doesn't have anything to do with the coordinate system. You get sines and cosines in your example because of nice boundary conditions.
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